Bạn tham khảo:
$7/$
$n_{NaOH}=\frac{200.2\%}{40}=0,1(mol)$
$\to n_{OH^{-}}=0,1(mol)$
$n_{Ba(OH)_2}=\frac{n_{OH^{-}}}{2}=0,05(mol)$
$V_{Ba(OH)_2}=\frac{0,05}{0,5}=0,1(l)$
$8/$
$CM_{HNO_3}=\frac{10.1,12.12,6}{63}=2,24M$
$HNO_3 \to H^{+} +NO_3^{-}$
$[H^{+}]=2,24M$
$9/$
BTĐT:
$2.0,4+0,5.2=x$
$\to x=1,8$
